Problem 111 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equations \(\log x+\log x^{1 / 2}+\log x^{1 / 4}\) \(+\ldots=y\) and \(\frac{1+3+5+\ldots . .+(2 y-1)}{4+7+10+\ldots .+(3 y+1)}\) \(=\frac{20}{7 \log x}\) is (A) \(x=10^{5}, 10^{-5 / 7}\) (B) \(y=10,-\frac{10}{7}\) (C) \(x=10,-\frac{10}{7}\) (D) \(y=10^{5}, 10^{-5 / 7}\)

Step-by-Step Solution

Answer

The solution is (A) \(x=10^{5}, 10^{-5/7}\).

Step 1: Simplify the First Equation

Consider the equation \(\log x + \log x^{1/2} + \log x^{1/4} + \ldots = y\). This is an infinite geometric series of the form \(\log x + \frac{1}{2}\log x + \frac{1}{4}\log x + \ldots \). The first term \(a = \log x\) and the common ratio \(r = \frac{1}{2}\). The sum \(S\) of the series is given by \( S = \frac{a}{1-r} = \frac{\log x}{1 - \frac{1}{2}} = 2 \log x \). Hence, \(y = 2 \log x\).

Step 2: Solve for \(x\) in Terms of \(y\)

From Step 1, we have \(y = 2 \log x\). Solving for \(x\), we get \(\log x = \frac{y}{2}\). Therefore, \(x = 10^{y/2}\).

Step 3: Evaluate the Second Equation

Given the equation \(\frac{1+3+5+\ldots +(2y-1)}{4+7+10+\ldots +(3y+1)} = \frac{20}{7 \log x}\), we first calculate the sum of these sequences. The numerator \(1+3+\ldots +(2y-1)\) is the sum of the first \(y\) odd numbers, which equals \(y^2\). The denominator \(4+7+10+\ldots +(3y+1)\) is an arithmetic series with first term \(4\), common difference \(3\), and \(y\) terms: sum = \(\frac{y}{2} (2\cdot4 + (y-1)3) = \frac{y}{2} (3y + 5)\).

Step 4: Substitute and Solve for \(y\)

Substituting these sums into the second equation gives: \(\frac{y^2}{\frac{y}{2} (3y + 5)} = \frac{20}{7 \log x}\). Solving for \(y\) in terms of \(\log x\), \(y^2 = \frac{y}{2} (3y + 5) \cdot \frac{20}{7 \log x}\). Express \(\log x\) using \(y = 2 \log x\) as \(\log x = \frac{y}{2}\). Substitute \(\log x\) back to get \(y = 10\).

Step 5: Verify Solutions

Verify the solutions using substitutions back into the equation. With \(y = 10\), \(\log x = \frac{y}{2} = 5\), so \(x = 10^{5}\). However, the requirement of balancing checks back on the equation also allows for \(x = 10^{-\frac{10}{7}}\). Therefore, the only correct form would balance as both \(x = 10^5\) and \(x = 10^{-5/7}\) are correct solutions.

Key Concepts

Infinite Geometric Series

An infinite geometric series is a sequence of numbers where each term is multiplied by a fixed, non-zero number called the common ratio to get the next term. In this series, the first term is labeled as \(a\), and the common ratio is \(r\). The series could go on indefinitely, but it has a convergent sum if \(|r| < 1\). This means you can actually find the sum of all its terms—even if there are infinitely many.

In the problem, \(\log x + \log x^{1/2} + \log x^{1/4} + \ldots\) forms an infinite geometric series. The first term \(a = \log x\) and the common ratio \(r = \frac{1}{2}\). When the common ratio is between \(-1\) and \(1\), the series can be summed up using the formula:

• \( S = \frac{a}{1 - r} \)

For this sequence, substituting \(a = \log x\) and \(r = \frac{1}{2}\), we get:

• \( S = \frac{\log x}{1 - \frac{1}{2}} = 2 \log x \)

Hence, in this case, \(y = 2 \log x\). This concept is an essential foundation for solving the equation in the exercise.

Arithmetic Sequence

An arithmetic sequence, also known as an arithmetic progression, is a series of numbers in which the difference between consecutive terms is constant. The general form of an arithmetic sequence is expressed as \(a, a + d, a + 2d, \ldots\), where \(a\) is the first term and \(d\) is the common difference between the terms.

In the exercise, the sequence \(4 + 7 + 10 + \ldots + (3y + 1)\) is an example of an arithmetic sequence. For this particular sequence, \(a = 4\) and the common difference \(d = 3\). To find the sum of the first \(n\) terms (\(S_n\)) of an arithmetic sequence, you can use the following formula:

• \( S_n = \frac{n}{2} (2a + (n-1)d) \)

Applying this to our sequence, the sum can be calculated as:

• \( \text{Sum} = \frac{y}{2} (2 \cdot 4 + (y - 1) \cdot 3) = \frac{y}{2} (3y + 5) \)

Understanding how to sum up the terms in an arithmetic sequence is crucial for working through the exercise correctly.

Logarithmic Equations

Logarithmic equations involve logarithms, which are the inverse operations of exponents. In simple terms, a logarithm tells you the power to which a number must be raised to obtain another number. In mathematical terms, if \(b^x = y\), then \(\log_b(y) = x\). Here, \(b\) is the base of the logarithm, and \(y\) is the value we want to find the power for.

The exercise requires understanding these equations due to terms like \(\log x\). When solving the equation \(y = 2 \log x\), we need to understand how to manipulate the logarithms. You can express \(x\) in terms of \(y\) as \(x = 10^{y/2}\) by rearranging the equation to isolate \(x\).

• Start with: \(y = 2 \log x\)
• Solve for \(\log x\): \(\log x = \frac{y}{2}\)
• Express \(x\) in exponential form: \(x = 10^{y/2}\)

Such manipulations show the versatility and usefulness of logarithmic equations in solving complex problems. It's vital to recognize that logarithms simplify the equations and help in finding solutions easily.